Problem: $\overline{AC}$ is $6$ units long $\overline{BC}$ is $9$ units long $\overline{AB}$ is $3\sqrt{13}$ units long What is $\tan(\angle ABC)$ ? $A$ $C$ $B$ $6$ $9$ $3\sqrt{13}$
Answer: SOH CAH TOA an = pposite over djacent opposite $= \overline{AC} = 6$ adjacent $= \overline{BC} = 9$ $\tan(\angle ABC)=\dfrac{6}{9}$ $=\dfrac{2}{3}$